1.TestA set of 4-in diameter direct shear tests were conducted on a lightly cementedsand. Test results are as shown. Plot the data (units = psf) and determine c' (inpsf) and '. How would c' change if there were no cementation?Number123Forces At FailureNormal(lb)60120180Shearing(lb)4071108Shear Stresses at FailureNormal (psf) Shearing (psf)688458137581420631238c' =c' =29.5 degrees57.3 psf0.4 psi

Given: Test Number Forces at failure (lb) Shear stresses at failure (psf) Normal Shearing Normal Shearing 1 60 40 688 458 2 120 71 1375 814 3 180 108 2063 1238 Soil sample = Cemented sand Diameter of the shear stress apparatus = 4 in To find: Plot the data Determine c' (psf) and ϕ' How would c' change if there were no cementation?Solution: For the direct shear test, after conducting at least 3 tests at different normal stresses, the maximum shear stresses derived at each one are plotted versus the normal stresses.For calculation of cohesion and angle of internal friction, the envelope yields the friction angle φ and cohesion c of the M-C Failure Criterion. From the graph, The cohesion of the cemented sand, c' = 58 psf = 5812×12=0.403 psi The angle of internal friction, ϕ'= 30o For no cemented soil, the cohesion of the soil, c' will be zero.Answer: c' = 58 psf ϕ'= 30o If there were no cementation, the cohesion of the soil becomes zero, c' = 0